A question on the rate of chemical reactions, a multiple choice question on the rate of chemical rea

1) Yes, as long as it is the substance in the equation.

2) The reaction rate ratio of O2 and SO2 is 11:8, so when the rate of SO2 generation is, the rate of O2 consumption is.

3) Within 4 seconds, the consumption of O2 is, and now the remaining O2 is, then O2 is at the beginning.

The reaction rate can be expressed as a change in the concentration of O2 or SO2.

- Fes2 or Fe2O3 cannot be used because they are solids and the concentration of solids is considered constant.

The rate of SO2 formation and the rate of O2 reduction are both rates of positive reactions.

- The ratio of the two rates is equal to the coefficient ratio.

-O2 reduction rate: mol·l-1·s-1*(11 8)=4s at the concentration of O2 decrease.

O2 at the beginning of the concentration.

mol·l-1+

Generally, the concentration or pressure change of the gas is selected to reflect the rate of chemical reaction.

The reaction rate ratio is the chemical coefficient ratio so it is 11:8, so when the rate of formation of SO2 is, the rate of consumption of O2 is.

This is directly asked according to the above question, if it is not impossible to calculate.

The rate of oxygen reduction calculated in the above problem is that it reacts for another 5 seconds, so the co-reaction is added, so it starts with 5

Units omitted).

1 The reaction rate indicates the speed of a chemical equation that can be used in the equation !! In addition to solids!! of substances.

2 The reaction rate ratio is the chemical coefficient ratio so is.

3 055*4= mol

But pay attention to the volume of the container).

1> reactants and products are acceptable, and the ratios expressed at each rate are the same as the ratios of each stoichiometry.

2> The ratio of the rate of SO2 formation to the rate of O2 reduction is 8:11, so the rate of O2 reduction v=(11 8) mol·l-1·s-1= mol·l-1·s-1

3> can be obtained from <2 >, and within 4 seconds, the reduced concentration of 02 is:

Then the concentration at the beginning of O2 is: (

Choose C. If the reactant is a pure solid or liquid, it is useless to increase the amount of reactant because the pure solid or liquid concentration is constant and does not affect the rate of the chemical reaction.

Increasing the pressure must result in a higher concentration in order to increase the reaction rate.

Increasing the temperature will increase the number of activated molecules, and the reaction rate will definitely be accelerated; Cooling will reduce the number of activated molecules, and the reaction rate will definitely slow down.

1.If the equivalent pressure is increased, the equilibrium shifts to the right, and the angle of the concentration quotient decreases and the equilibrium shifts to the right. Because of the increase of concentration, the rate of the forward and negative reactions increases, but the rate of the positive reaction increases more.

236 According to Le Chatre's principle, we know that the conditions for cooling, pressurizing, and removing the product should be given. 3

Choosing 135 can first calculate that some c is generated, so some a, and b participate in the reaction. So you can know that 1 is right, 2 is wrong, and the conversion rate of A and B is 40%. 3 is right.

At equilibrium, the concentration of b should be the same, and the coefficients of the reaction formula are the same, equivalent equilibrium, 5 pairs.

After 5 min, c(a)=3xmol l, c(b)=5xmol l, before the reaction, c(a)=(3x

c(b)=(5x

Yes. 3xx = 1) at this point a concentration.

c(a)=3xmol/l=

The pure standby concentration of a before the reaction.

c(a)=(3x

The amount of substances a, b before the reaction.

n(b)=n(a)=

2l=3mol

2) B's sail is flat and the leg is responsive.

v(b)=3)v(c)=2v(b),x=2

1. Yes. Because it is a closed container rather than a container with a certain volume, the pressure is increased, that is, the volume decreases, the concentration of reactants increases, the number of activated molecules per unit volume increases, the number of effective collisions per unit time increases, and the reaction rate accelerates. Adding charcoal doesn't seem to be able to speed up.

2. No, you can't. Because the volume of the container does not change, the actual concentration does not change, and the reaction rate does not change.

3. No, you can't. Because chloride ions are not involved in the reaction.

Pressure conditions. For chemical reactions with gas participation, when other conditions remain unchanged (except volume), the pressure is increased, that is, the volume decreases, the concentration of reactants increases, the number of activated molecules per unit volume increases, the number of effective collisions per unit time increases, and the reaction rate accelerates. Otherwise, it decreases. If the volume is constant, the rate of the pressurized reaction (adding gases that do not participate in this chemical reaction) will not change.

Because the concentration does not change, the number of activated molecules per unit volume does not change. However, when the volume remains the same, the reactants are added, which is also pressurized, and the concentration of reactants is increased, and the rate will also increase.

Temperature conditions. As long as the temperature is raised, the reactant molecules gain energy, so that some of the molecules with lower energy become activated molecules, which increases the percentage of activated molecules, so that the number of effective collisions increases, so the reaction rate increases (the main reason). Of course, as the temperature increases, the rate of molecular motion accelerates, and the reaction will increase as the number of molecular collisions of reactants per unit time increases

Catalyst. The use of positive catalysts can reduce the energy required for the reaction, so that more reactant molecules become activated molecules, and the percentage of reactant molecules per unit volume is greatly increased, thereby increasing the reactant rate thousands of times. Negative catalysts are the opposite.

Catalysts can only change the rate of chemical reactions, but not the equilibrium of chemical reactions.

Conditional concentration. When other conditions are the same, increasing the concentration of reactants increases the number of activated molecules per unit volume, thereby increasing the effective collision and increasing the reaction rate, but the percentage of activated molecules remains the same.

Solution: 3a +b = 2c +xd start 6 5 0 0 reaction 3a a 2a xa equilibrium 6-3a 5-a 2 5 so a = 1, x = 3 4

Conversion rate of b = 1 5 100%.

Equilibrium a concentration = (6-3) 4=

The following chemical reactions are more chaotic (s).

BASO4+CO32-= (reversible) BACO3+SO42-

1) According to the above transformation equation, it can be seen that K=C(SO42-) C(CO32-)=KSP(BaSO4) KSP(BAC3)=When paying attention to equilibrium, C(Ba2+), C(SO42-) and C(CO32-) must satisfy both the dissolution equilibrium constant of BaSO4 and the dissolution equilibrium constant of BaCO3, so the above equation holds. ）

2) After the complete transformation of Baso4, C(SO42-) = in the solution, according to the equilibrium constant just calculated, the concentration of CO32- in the solution can be calculated, that is: K = C(SO42-) C(CO32-) =, C (CO32-) = in the solution can be obtained

The reactive ion equation is: BASO4+CO3(2-)====BAC3+SO4(2-) is the combination of the two equations.

1) The equilibrium constant k=c(SO4(2-)) c(CO3(2-)) is equivalent to dividing the KSP of BASO4 by BACO3.

2) When completely dissolved, the sulfate concentration is calculated according to the equilibrium constant calculated in the previous question.

Ideas: This question organically combines the chemical reaction rate, chemical equilibrium and time rate change diagram, and uses the knowledge of number combinations, such as time and displacement images in uniform motion in physics.

Analysis idea: The shaded area in the figure is equivalent to v positive * t v inverse * t, since v in this problem is the reaction rate expressed by the change in the concentration of b, so v positive * t v inverse *t is equivalent to the difference between the concentration of b reacted in the reaction and the concentration of b generated, that is, the net decrease in b concentration. However, among the four options provided, there is no "reduction in b concentration" option, but from the equation:

B is the reactant and C is the product, and the coefficient of the two is the same, and the decrease of B concentration is the increase of C concentration. So the answer is c

Brief comment: The difficulty in solving this question is whether the candidate can understand the important information that "v positive * t v inverse * t is equivalent to the difference between the concentration of b reacted in the reaction and the concentration of b produced, that is, the net decrease in the concentration of b". To understand the above information, we must use the time and displacement images in uniform motion in physics to understand why the shadow area is equivalent to v positive *t v inverse *t!

However, I think that this problem should be understood with the help of time and displacement images in uniform motion in physics to understand that the shadow area is equivalent to v positive *t v inverse *t, which is really far-fetched to say that it is an organic combination! Because the image of uniform motion in physics is similar to the image of displacement in this image, but not exactly the same, I think few students can do it right if students are asked to make a big turn from a chemical perspective and then dig out this information that is not relative to each other. There are only a few right ones, and most of them are the right ones!

Therefore, I boldly speculate that the author of this question is really a failure, and I hope that this kind of question will not appear again in future practice problems!

I think the C selection process is a bit cumbersome.